Question

In the decomposition of limestone to lime, the values of \( \Delta H^\circ \) and \( \Delta S^\circ \) are \( +179.1 \, kJ \, mol^{-1} \) and \( 160.2 \, J \, K^{-1} \, mol^{-1} \). Calculate the temperature above which conversion of limestone to lime will be spontaneous, if values of \( \Delta H^\circ \) and \( \Delta S^\circ \) remain unchanged with temperature. [Assuming pressure 1 bar]

• A. 1018 K
• B. 1200 K
• C. 1028 K
• D. 1118 K

Hint:

For temperature-based spontaneity problems, use \( \Delta G = 0 \) conditionAlways convert \( \Delta H \) into Joules to match units of \( \Delta S \).

Answer 1

The Correct Option is D

Step 1: Use Gibbs free energy condition.
For spontaneity:
\[ \Delta G = \Delta H - T\Delta S \]
At equilibrium (boundary of spontaneity):
\[ \Delta G = 0 \]
Thus:
\[ \Delta H = T\Delta S \]

Step 2: Convert units properly.
\[ \Delta H = 179.1 \, kJ \, mol^{-1} = 179100 \, J \, mol^{-1} \]
\[ \Delta S = 160.2 \, J \, K^{-1} \, mol^{-1} \]

Step 3: Calculate temperature.
\[ T = \frac{\Delta H}{\Delta S} \]
\[ T = \frac{179100}{160.2} \]
\[ T \approx 1118 \, K \]

Step 4: Interpretation.
For \( T > 1118 \, K \), \( \Delta G < 0 \), so the reaction becomes spontaneous.

Step 5: Conclusion.
Thus, the temperature above which the reaction is spontaneous is:
\[ \boxed{1118 \, K} \]